1. CHAPTER 2: STATICS OF PARTICLES
STATICS AND DYNAMICS
PDT 101/3
CHAPTER 2: STATICS OF PARTICLES
PREPARED BY:
TAN SOO JIN

2. CHAPTER OUTLINE Scalars and Vectors Vector Operations
Vector Addition of Forces
Addition of a System of Coplanar Forces

3. APPLICATION OF VECTOR ADDITION
There are three concurrent forces acting on the hook due to the chains.
We need to decide if the hook will fail (bend or break).
To do this, we need to know the resultant or total force acting on the hook as a result of the three chains. FR

4. 2.1 Scalars and Vectors Scalar
– any positive or negative physical quantity that can be completely specified by its magnitude.
e.g. Mass, time, and length

5. 2.1 Scalars and Vectors Vector
– A quantity that has magnitude and direction
e.g. Position, force and moment
– Represent by a letter with an arrow over it,
– Magnitude is designated as
– In this subject, vector is presented as A and its magnitude (positive quantity) as A

6. Scalar Multiplication
2.2 Vector Operations
Vector A and its
negative counterpart
Scalar Multiplication
and Division

7. 2.2 Vector Operations Vector Addition Parallelogram Law
Triangle construction
Resultant vector R can be found by triangle construction

8. 2.2 Vector Operations Vector Subtraction or A - B
R’ = A – B = A + (-B)

9. 2.3 Vector Addition of Forces
Finding a Resultant Force
Parallelogram law is carried out to find the resultant force
Resultant,
FR = (F1+ F2)

10. a/sin A = b/sin B = c/sin CC
Law of Cosines
Law of Sinus
a2 = b2 + c2 – 2bc(cosA)
b2 = a2 + c2 – 2ac(cosB)
c2 = a2 + b2 – 2ab(cosC)
a/sin A = b/sin B = c/sin CC

11. 2.3 Vector Addition of Forces
Procedure for Analysis
Parallelogram Law
Make a sketch using the parallelogram law
2 components forces add to form the resultant force
Resultant force is shown by the diagonal of the parallelogram
The components is shown by the sides of the parallelogram

12. 2.3 Vector Addition of Forces
Procedure for Analysis
Trigonometry
Redraw half portion of the parallelogram
Magnitude of the resultant force can be determined by the law of cosines
Direction if the resultant force can be determined by the law of sines
Magnitude of the two components can be determined by the law of sines

13. EXAMPLE 1
The screw eye is subjected to two forces, F1 and F2. Determine the magnitude and direction of the resultant force.

14. SOLUTION
Parallelogram Law
Unknown: magnitude of FR and angle θ

15. How to solve the question?

16. Trigonometry
Law of Cosines
Law of Sines

17. Trigonometry
Direction Φ of FR measured from the horizontal

18. EXERCISE 1
If the tension in the cable is 400 N, determine the magnitude and direction of the resultant force acting on the pulley. This angle is the same angle of line AB on the tailboard block. (Ans: 400N, 60o)

19. 2.4 Addition of a System of Coplanar Forces
Scalar Notation
- x and y axes are designated positive and negative
- Components of forces expressed as algebraic scalars

20. 2.4 Addition of a System of Coplanar Forces
Cartesian Vector Notation
Cartesian unit vectors i and j are used to designate the x and y directions
Unit vectors i and j have dimensionless magnitude of
unity ( = 1 )
Magnitude is always a positive quantity, represented by scalars Fx and Fy

21. Coplanar Force Resultants
To determine resultant of several coplanar forces:
- Resolve force into x and y components
- Addition of the respective components using scalar
algebra
- Resultant force is found using the parallelogram law
- Cartesian vector notation:

22. 2.4 Addition of a System of Coplanar Forces
Coplanar Force Resultants
- Vector resultant is therefore
- If scalar notation are used

23. 2.4 Addition of a System of Coplanar Forces
Coplanar Force Resultants
-In all cases we have
* Take note of sign conventions
- Magnitude of FR can be found by Pythagorean Theorem

24. EXAMPLE 2
Resolve the 500 N force exerted on the hook in Figure into horizontal and vertical components.

25. SOLUTION 2

26. EXERCISE 2
Determine the x and y components of the forces shown in Figure below.
Fx = -200 kN
Fy = 210 kN
Fx = kN
Fy = kN

27. EXAMPLE 3
Determine x and y components of F1 and F2 acting on the boom. Express each force as a Cartesian vector.

28. SOLUTION
Scalar Notation
Hence, from the slope triangle, we have

29. By similar triangles we have
Scalar Notation:
Cartesian Vector Notation:

30. EXERCISE 4
Determine the x and y components of the forces of the two forces F1 and F2 acting on the eye-bolt shown in Figure below.
(Ans: F1x=173.21N, F1y=100N
F2x=-76.6N, F2y= 64.28N)

31. EXAMPLE 4
The link is subjected to two forces F1 and F2. Determine the magnitude and orientation of the resultant force.

32. SOLUTION I
Scalar Notation:

33. Resultant Force
From vector addition, direction angle θ is

34. SOLUTION II Cartesian Vector Notation
F1 = { 600cos30°i + 600sin30°j } N
F2 = { -400sin45°i + 400cos45°j } N
Thus,
FR = F1 + F2
= (600cos30ºN - 400sin45ºN)i (600sin30ºN + 400cos45ºN)j
= {236.8i j}N
The magnitude and direction of FR are determined in the same manner as before.

35. EXERCISE 3
Find the x and y components of the forces P and Q shown in Figure below. Calculate magnitude and direction of the resultant force.
(Ans: FR= 40.26kN, θ= 75.62o)

36. EXERCISE 4
If Φ=30° and the resultant force acting on the gusset plate is directed along the positive x-axis, determine the magnitudes of F2 and the resultant force.
(Ans: F2=12.93 kN, FR=13.19 kN)